Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $t \neq 0$. $z = \dfrac{t^2 + 2t - 24}{t - 9} \div \dfrac{t + 6}{-6t + 54} $
Dividing by an expression is the same as multiplying by its inverse. $z = \dfrac{t^2 + 2t - 24}{t - 9} \times \dfrac{-6t + 54}{t + 6} $ First factor the quadratic. $z = \dfrac{(t + 6)(t - 4)}{t - 9} \times \dfrac{-6t + 54}{t + 6} $ Then factor out any other terms. $z = \dfrac{(t + 6)(t - 4)}{t - 9} \times \dfrac{-6(t - 9)}{t + 6} $ Then multiply the two numerators and multiply the two denominators. $z = \dfrac{ (t + 6)(t - 4) \times -6(t - 9) } { (t - 9) \times (t + 6) } $ $z = \dfrac{ -6(t + 6)(t - 4)(t - 9)}{ (t - 9)(t + 6)} $ Notice that $(t - 9)$ and $(t + 6)$ appear in both the numerator and denominator so we can cancel them. $z = \dfrac{ -6\cancel{(t + 6)}(t - 4)(t - 9)}{ (t - 9)\cancel{(t + 6)}} $ We are dividing by $t + 6$ , so $t + 6 \neq 0$ Therefore, $t \neq -6$ $z = \dfrac{ -6\cancel{(t + 6)}(t - 4)\cancel{(t - 9)}}{ \cancel{(t - 9)}\cancel{(t + 6)}} $ We are dividing by $t - 9$ , so $t - 9 \neq 0$ Therefore, $t \neq 9$ $z = -6(t - 4) ; \space t \neq -6 ; \space t \neq 9 $